Answer:
Step-by-step explanation:
4. h(x) = 2x^2 + 14x - 60
Step-by-step explanation:
Given that h(x) is a quadratic.
Also, h(3) = h(-10) = 0
(A) h(x) = x^2 - 13x - 30
=> h(3) = 3^2 - 13(3) - 30
=> h(3) = 9 - 39 - 30
=> h(3) = -30 - 30
=> h(3) = -60
=> h(3) ≠ 0
(B) h(x) = x^2 - 7x - 30
=> h(3) = 3^2 - 7(3) - 30
=> h(3) = 9 - 21 - 30
=> h(3) = -12 - 30
=> h(3) = -42
=> h(3) ≠ 0
(C) h(x) = 2x^2 + 26x - 60
=> h(3) = 2(3^2) + 26(3) - 60
=> h(3) = 2(9) + 78 - 60
=> h(3) = 18 + 78 - 60
=> h(3) = 96 - 60
=> h(3) = 36
=> h(3) ≠ 0
(D) h(x) = 2x^2 + 14x - 60
=> h(3) = 2(3^2) + 14(3) - 60
=> h(3) = 2(9) + 42 - 60
=> h(3) = 18 + 42 - 60
=> h(3) = 60 - 60
=> h(3) = 0
And
=> h(-10) = 2(-10)^2 + 14(-10) - 60
=> h(-10) = 2(100) - 140 - 60
=> h(-10) = 200 - 200
=> h(-10) = 0
Clearly we have,
=> h(3) = h(-10) = 0
Hence, the correct option is (D) h(x) = 2x^2 + 14x - 60
Answer:
N÷4×K
Step-by-step explanation:
It would be easy to explain if the unknown numbers had values
M=2 do need a step by step explanation?
Answer:
- x ≥ 2009
- 3.2x -6222.8
- 257.2 mm
- year 2100
Step-by-step explanation:
1. The problem statement tells you the function applies for year values (x) 2009 and later. The domain is real numbers greater than or equal to 2009.
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2. We can use the distributive property to eliminate parentheses:
3.2(x -2009) +206 = 3.2x -6428.8 +206
= 3.2x -6222.8
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3. Put 2025 in the equation and do the arithmetic
g(2025) = 3.2·2025 -6222.8 = 257.2 . . . . mm
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4. Put this value of level rise in the equation and solve for x.
g(x) = 3.2x -6222.8
500 = 3.2x -6222.8 . . put 500 mm where g(x) is in the equation
6722.8 = 3.2x . . . . . . . add 6222.8
x = 6722.8/3.2 = 2100.875
The water rise will be equal to about half a meter late in the year 2100.
8.3333333333333333333333333333333333333 meters I recommend putting 8.3 with a bar notation over the 3